Mechanical vibrations theory and applications 1st edition kelly solutions manual

Mechanical Vibrations Theory and Applications 1st Edition Kelly Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/mechanical-vibrations-theory-and-applications-1st-edition-kelly-solutions-manu al/

CHAPTER 6: TWO DEGREE-OFFREEDOM SYSTEMS Short Answer Problems 6.1 True: A two-degree-of-freedom system has two natural frequencies. The normal mode solution is assumed and substituted into the differential equations. Two linearly independent solutions are obtained, one corresponding to each natural frequency.

6.2 False: The natural frequencies are determined by setting |

|

0.

6.3 False: The natural frequencies of a two degree-of-freedom system are independent of the choice of generalized coordinates used to model the system.

6.4 True: Even powers occur if the system is undamped. If the system has viscous damping odd powers occur as well.

6.5 False: The modal fraction represents the displacement of the second generalized coordinate compared to the displacement of the first.

6.6 False: The principal coordinates are the generalized coordinates for which the mass matrix and the stiffness matrix are diagonal matrices.

6.7 False: The free response of a damped two degree-of-freedom system has two modes of vibration, both of which may be underdamped.

6.8 True: A displacement of a node for a mode of a two degree-of-freedom system can serve as a principal coordinate. Since a node can only occur for the second mode, the node can serve as a principal coordinate for the first mode.

6.9 True: While the natural frequencies are independent of the choice of generalized coordinates, the modal fractions for a two degree-of-freedom system depend upon the choice of generalized coordinates.

6.10 True: The sinusoidal transfer function which is the transfer function evaluated at , where is the frequency of input, can be used to determine the steady-state response of a . two-degree-of-freedom system. The steady-state response is 6.11 True: The addition of a vibration absorber adds another degree of freedom to the original system.

6.12 False: The undamped vibration absorber is tuned to the excitation frequency to eliminate steady-state vibrations of the primary systems.

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Chapter 6: Two Degree-of-Freedom Systems 6.13 False: An optimally tuned damped vibration absorber is tuned such that the amplitude of vibration during start up is minimized as well as increasing the operating range.

6.14 False: Addition of a dynamic vibration absorber to a damped primary system will not eliminate the steady-state vibrations of the primary system at any speed.

6.15 True: A Houdaille vibration absorber is used on engine crankshafts and other devices.

6.16 6.17 . It is substituted into the differential 6.18 The normal mode solution is equations for free vibration of a two-degree-of-freedom system to determine the natural | 0 and frequencies and mode shapes. The mode shapes are the solutions to | the mode shapes are the corresponding non-trivial solutions of KX= MX. while for a damped 6.19 The assumed solution for an undamped system is system . The values of for a damped system may be complex while for an undamped system must real. The modal fraction for a damped system are also complex and occur in complex conjugate pairs.

6.20 A real solution for overdamped.

means that the mode associated with the value of

is

6.21 A complex solution for means that the mode associated with the value of is underdamped. Also the complex conjugate of the value is also a solution of the fourth order equation.

6.22 The transfer function , is the transform of the response at the generalized due to a unit impulse applied at the generalized coordinate coordinate described by whose response is described by .

6.23 The sinusoidal transfer function is where is the frequency of excitation. It can | is the amplitude of the steady-state response | where | be written as | and is the phase angle between the excitation and the response.

6.24 The differential equations defining the principal coordinates are 0 0 6.25 A modal fraction equal to zero implies that the second mode is uncoupled from the first mode.

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Chapter 6: Two Degree-of-Freedom Systems 6.26 A modal fraction equal to one means that the system moves as a rigid body for that mode.

6.27 There are zero nodes corresponding to the lowest natural frequency of a tow degree of freedom system.

6.28 If the differential equations governing a two degree-of-freedom system are uncoupled when a certain set of generalized coordinates are used the coordinates must be principal coordinates for the system.

6.29 (b) the denominator D(s) 6.30 The appropriate convolution integral solution is

where

.

6.31 The amplitudes and phases are determined by applying the initial conditions to sin sin 6.32

|

|

where |

|

and

tan 6.33 The vibration amplitude of the primary system when a dynamic vibration absorber tuned to the excitation frequency is added to the system is zero.

6.34 A dynamic vibration absorber works by adding one degree of freedom to a system whose natural frequency is ear the excitation frequency such that the two natural frequencies are away from the excitation frequency. One natural frequency is lower than the excitation frequency one is higher.

6.35 A vibration damper is used in situations where vibration control is necessary over a wide range of frequencies.

6.36 The two problems addressed by adding damping to a vibration absorber are the large amplitude during start up when the lowest natural frequency of the system is passed and the wide variation in amplitude of the primary system with operating speed.

6.37 The optimum damping ratio for a Houdaillle damper is

where

is

the ratio of the moment of inertia of the damper to the moment of inertia of the shaft.

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Chapter 6: Two Degree-of-Freedom Systems 6.38 Given: 6

27

21

0. The natural frequencies are obtained by applying the 3.5, 1 leading to

quadratic formula 6.39 Given:

3

1

2 2 |

2 for a system such that | 1 0

1.87,1.

0 (a) The equation is the frequency equation 0 0. Thus the differential equations are

0 1

3 2

0 0

2 2

which comes from the system shown in Fig SP 6.39a. (b) The natural frequencies of the 3 2 0 system are calculated from 3 2 4 2 2 √ 5 2 0whose solutions are 0.662, 2.14 (c) The mode which

shapes are determined from the given equation. The first equation yields 1.28 and

leads to

1 and 1.28

0.781 which lead to mode shapes of

1 (e) The mode shapes are illustrated in Figure SP 6.39b.

0.781

6.40 The mode shape corresponds to the higher mode. (a)

6.41 Given:

3

= 0.0120

(b) The squares of the natural frequencies are the roots of and 2. Hence = 1 and √2 1.41 6.42 Given: given by | 4 | tan

sin 4 2.5

2

0 which are 1

2.5 sin 4 The steady-state response is

, where

0.3853. Thus

3

0.00274

|10 0.0107 sin 4

9.89

4.01i |

0.007 and

0.3853 .

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Chapter 6: Two Degree-of-Freedom Systems 1 (b)

6.43 (a) 6.44 Given: ,

1

√

81.5

1.05 (c) 100,

0.2

1.05 1

0.2

1.05

1.05 1

2 0.2

0.2

1 1.05

1

, 128.8

6.45 Given: f = 30 Hz, 200 N (a) The steady-state vibration of the machine are eliminated if the absorber is tuned to a frequency of 30 Hz = 188.5 rad/s (b) If 3 kg the stiffness of the absorber is 3 kg 188.5 rad/s 1.07 10 N/m (c) 1.9 mm (d) 188.5 rad/s N/

.

50 kg,

6.46 Given: .

0.833,

10 kg,

100 rad/s (a)

0.2,

0.833 100

83.33 rad/s (b)

. .

0.25.

6.47 Given:

0.1 kg · m ,

(a)

0.1,

0.002

. .

0.01 kg · m ,

0.002 (a)

.

.

0.465

0.1, (b)

Θ

0.042 rad

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Chapter 6: Two Degree-of-Freedom Systems Chapter Problems 6.1 Derive the differential equation governing the two degree-of-freedom system shown in Figure P6.1 using and as generalized coordinates.

Given: System shown Find: Differential equations Solution: Free-body diagrams of both masses are drawn at an arbitrary instant

Summation of forces on the FBD’s lead to

and 2

2

The equations are rearranged to 2

0

2

3

0

Problem 6.1 illustrates the use of Newton’s law to derive the differential equations for a 2DOF system.

6.2 Derive the differential equation governing the two degree-of-freedom system shown in Figure P6.2 using x and as generalized coordinates.

Given: System shown Find: Differential equations Solution: Free-body diagrams of the bar and the discrete mass, assuming small , are drawn at an arbitrary instant 429 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems

Summing moments on the bar using ∑ 2

∑ sin

2

Summing forces on the mass using ∑

∑

leads to 2

2

1 12

leads to

2

Rearranging Eqs. (a) and (b) yields 1 3

3 4

2

sin

0

2

Problem 6.2 illustrates the use of D’Alembert’s principle to derive the differential equations governing a 2DOF system.

6.3 Derive the differential equations governing the two degree-of-freedom system shown in Figure P6.3 using and as generalized coordinates.

Given: System shown Find: Differential equations Solution: Free-body diagrams of the disks are drawn at an arbitrary instant

430 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems

Summing moments on disk A using ∑

∑

leads to

Summing moments on disk B using ∑

∑

leads to

sin

2

2

2

Rearranging Eqs. (a) and (b) lead to 2

0

and 4

2

sin

Problem 6.3 illustrates the use of D’Alembert’s principle to derive the differential equations governing a 2DOF system.

6.4 Derive the differential equations governing the two degree-of-freedom system shown in and as generalized Figure P6.4 using coordinates.

Given: System shown Find: Differential equations 431 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems Solution: Free-body diagrams of the disks are drawn at an arbitrary instant

Summing moments on disk A using ∑

∑

leads to

Summing moments on disk B using ∑

∑

leads to

Noting that 3 /2

Eqs. (a) and (b) are rearranged to 3 /2

3 /2

3 /2

3 /2

0

and 0

Problem 6.4 illustrates the use of D’Alembert’s principle to derive the differential equations governing a 2DOF system.

6.5 A two degree-of-freedom model of an airfoil shown in Figure P6.5 is used for flutter analysis.

Derive the governing differential equations using h and as generalized coordinates.

Given: , Find: Differential equations Solutions: FBD’s of the airfoil at an arbitrary instant are drawn.

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Chapter 6: Two Degree-of-Freedom Systems

Summing Forces on the FBD using ∑

∑

leads to

ℓ

Summing moments according to ∑

∑ ℓ

leads to ℓ

Rearranging the equations leads to ℓ

0 ℓ

ℓ

0

Problem 6.5 illustrates the derivation of differential equations governing the motion of a 2DOF system.

6.6 Derive the differential equations governing the damped two degree-ofand as freedom system shown using generalized coordinates.

Given: System shown Find: Differential equations Solution: Free-body diagrams of the blocks are drawn at an arbitrary instant

Summing forces on the first FBD leads to 2

2

2

Summing forces on the second FBD leads to sin 433 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems Equations (a) and (b) are rearranged to 2

3

3

0

and sin

Problem 6.6 illustrates the use of Newton’s law to derive the differential equations for a 2DOF system with viscous damping.

6.7 Derive the differential equations governing the damped two degree of freedom system shown in Figure P6.7 using and as generalized coordinates.

Given: System shown Find: Differential equations Solution: Free-body diagrams of the blocks are drawn at an arbitrary instant

Summing forces on the first FBD leads to

Summing forces on the second FBD leads to 2

Equations (a) and (b) are rearranged to 2

0

and 434 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems 0

2

Problem 6.7 illustrates the use of Newton’s law to derive the differential equations for a 2DOF system with viscous damping.

6.8 A two degree-of-freedom model of a machine as tool is illustrated in Figure P6.8. Using and generalized coordinates, derive the differential equations governing the motion of the system.

Given: System shown Find: Differential equations

Solution: Free-body diagram of the bar is drawn at an arbitrary instant

Summing moments on the FBD using ∑

∑

leads to

Summing moments on the FBD using ∑

∑

leads to

Rearranging Eqs. (a) and (b) leads to 0

and 0

Problem 6.8 illustrates the use of D’Alembert’s principle to derive the differential equations governing a 2DOF system with viscous damping.

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Chapter 6: Two Degree-of-Freedom Systems

6.9 Derive the differential equation of the two degree-of-freedom model of the machine tool of Chapter Problem 6.8 using x and as generalized coordinates.

Given: System shown Find: Differential equations

Solution: Free-body diagram of the bar is drawn at an arbitrary instant

Summing moments on the FBD using ∑

∑

leads to

Summing moments on the FBD using ∑

∑

leads to

Rearranging Eqs. (a) and (b) leads to 0

and 0

Problem 6.9 illustrates the use of D’Alembert’s principle to derive the differential equations governing a 2DOF system with viscous damping.

6.10 Determine the natural frequencies of the system of Figure P6.1 if m = 10 kg and k = 1 × 105 N/m. Determine and graphically illustrate the mode shapes.

Identify any nodes.

Given: m = 10 kg, k = 1 × 105 N/m 436 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems Find:

,

,

,

Solution: The differential equations are derived in the solution of Chapter Problem 6.1.

They are written in matrix form using the given values of k and m as 10 0

0 20

2 1

10

1 3

0 0

The natural frequencies are determined as the solutions of | 10

2 20 200

where

0

3 70

10

|

2

20

0 or 3

5

10 . The solutions of the equation are 100

rad s

158.11

rad s

The modal fractions are determined by 10

which when evaluated lead to

1 and

2

0.5. Hence the mode shapes are 1 0.5

1 2

The mode shape vectors are illustrated below. There is a node for the second mode in the spring connecting the two masses.

Problem 6.10 illustrates the computation of natural frequencies and mode shapes for 2DOF systems.

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Chapter 6: Two Degree-of-Freedom Systems 6.11 Determine the natural frequencies of the system of Figure P6.2 if m = 2 kg, L = 1 m and k = 1000 N/m.

Determine the modal fractions for each mode.

Given: m = 2 kg, L = 1m, k = 1000 N/m Find:

,

,

,

Solution: The differential equations are derived in the solution of Chapter Problem 6.2 as 3 4

1 3

0

2 0

2

Rewriting Eqs. (a) and (b) in matrix form using the given values 0.667 0

0 2

750 500

0 0

500 1000

The natural frequencies are determined as the solutions of | 750 500 0.667 0 1000 500 2 1.33 2166.7 500000

0.667

| 750

2

0 or 1000

500

The solutions of the equation are 16.69

rad s

36.69

rad s

The modal fractions are determined by 0.667 500

which when evaluated lead to

1.128 and 1 1.128

750

0.2949. Hence the mode shapes are 1 0.2949

Problem 6.11 illustrates the computation of natural frequencies and mode shapes for 2DOF systems.

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Chapter 6: Two Degree-of-Freedom Systems 6.12 Determine the natural frequencies of the system of Figure P6.3 if m = 30 g, I1 = 8 × 10-6 kg · m2, I2 = 2 × 10-5 kg · m2, r = 5 mm, and k = 10 N/m.

Determine the modal fraction for each mode.

8

Given: m = 30 g, 2

10

10

kg · m ,

kg · m , r = 5 mm, k = 10 N/m

Given: m = 2 kg, L = 1 m, k = 1000 N/m Find:

,

,

,

Solution: The differential equations are derived in the solution of Chapter Problem 6.3 as 2

0

4

2

sin

Rewriting Eqs. (a) and (b) in matrix form using the given values 10

0.875 0

0 2.3

5 2.5

10

0 0

2.5 2.5

The natural frequencies are determined as the solutions of | 0.875 2.5

where

10

and

5

2.5 2.5 5

2.3 0.875

10

|

0 or

0 2.3

2.5

2.5

The solutions of the equation are 2.219

rad s

7.943

rad s

The modal fractions are determined by 0.0875 2.5

which when evaluated lead to

1.827 and

1 1.827

5

0.2082. Hence the mode shapes are 1 0.2082

Problem 6.12 illustrates the computation of natural frequencies and mode shapes for 2DOF systems.

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Chapter 6: Two Degree-of-Freedom Systems 6.13 Determine the natural frequencies of the system 0.3 kg · m , 0.4 kg · m , of Figure P6.4 if N 1.6 10 m , 80 10 , and L = 30 cm. Determine the modal fractions for each mode. Identify any nodes.

Given: 1.6

0.3 kg · m , 10 m ,

0.4 kg · m , 80 10

N

, and L = 30 cm.

Find: , , , Solution: The differential equations are derived in the solution of Chapter Problem 6.4 as 3 /2

0

3 /2

and 3 /2

0

3 /2

Rewriting Eqs. (a) and (b) in matrix form using the given values 0.3 0 0 0.4

7.111 2.844

10

2.844 2.844

0 0

The natural frequencies are determined as the solutions of | 0.3

where

7.111 2.844

0.4 0.3

2.844 2.844 7.111

|

0 or

0 0.4

2.844

2.844

10 . The solutions of the equation are 61.11

rad s

164.6

rad s

The modal fractions are determined by 0.3

which when evaluated lead to

7.111 2.844

2.106 and 1 2.106

0.358. Hence the mode shapes are 1 0.358

A node exists for the second mode in the shaft that connects the two disks.

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Chapter 6: Two Degree-of-Freedom Systems Problem 6.13 illustrates the computation of natural frequencies and mode shapes for 2DOF systems.

6.14 An overhead crane is modeled as a two degreeof-freedom system as shown in Figure P6.14. The crane is modeled as a mass of 1000 kg on a steel (E = 200 × 109 N/m2) fixed-fixed beam with a moment of inertia of 4.2 × 10-3 m4 and length of 12 m.

The crane has an elastic rope of diameter 20 cm. At a specific instant, the length of the rope is 10 m and is carrying a 300 kg load. What are the two natural frequencies of the system?

Given: = 1000 kg, E 200 10 m, 300 kg Find:

,

10

N

, I = 4.2

10

m , L = 12 m, r = 10 cm,

10

m

,

Solution: The stiffness of the beam is 192

192 200

N 4.2 m 12 m

10

9.33

10

N m

The stiffness of the cable is 200

10

N π 0.1 m m 10 m

5.24

10

N m

The differential equations for a model of the system as two suspended masses is 0

0 0

0

or 1000 0

0 300

6.16 5.24

10

The natural frequencies are the solutions of | 266.6

rad s

0 0

5.24 5.24

|

0 which are

1513.7

rad s

Problem 6.14 illustrates the modeling and natural frequency computation of a 2DOF system.

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Chapter 6: Two Degree-of-Freedom Systems

6.15 A seismometer of mass 30 g and stiffness 40 N/m is used to measure the vibrations of a SDOF system of mass 60 g and natural frequency 150 rad/s. It is feared that the mass of the seismometer may affect the vibrations that are to be measured. Check this out by calculating the natural frequencies of the two degree-of-freedom with the seismometer attached.

60 g,

Given: Find:

,

150

,

30 g,

40 N/m

,

Solution: The stiffness of the original system is 0.06 kg 150 rad/s

1350 N/m

The 2DOF model of the seismometer attached to the machine is that of a mass suspended by a spring suspended from a mass-spring system. The governing differential equations for the model are 0

0 0

0

or 0.06 0

0 0.03

1390 40

40 40

The natural frequencies are the solutions of | 35.95

rad s

0 0

|

0 which are

152.3

rad s

Problem 6.15 illustrates the modeling and natural frequency computation of a 2DOF system.

6.16 Calculate the natural frequencies and modal fractions for the system of Figure P6.16.

Given: System shown Find:

,

,

,

Solution: FBD’s of the system at an arbitrary instant are shown below

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Chapter 6: Two Degree-of-Freedom Systems

Summing forces on the FBD’s using ∑

∑

leads to

3

6000

4000

0

4

4000

4000

0

Assuming a normal mode solution to the differential equations yields 3

6000

4000

0

4

4000

4000

0

and

These equations have a non-trivial solution if and only if 3

6000 4000

4

4000 4000

4000

4000

0

Evaluation of the determinant leads to 3

6000

4

4000

0

or 12

38000

1.8 10

15.54

rad s

ω

0

The solutions are ω

52.52 rad/s

The modal fractions are determined from 3

6000 4000

which leads to 1 1.388

1 0.569 443

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Chapter 6: Two Degree-of-Freedom Systems Problem 6.16 illustrates natural frequencies and mode shapes of a 2DOF system.

6.17 Determine the forced response to the system of Figure P6.1 and Chapter Problems 6.1 and 6.10 if the left-hand mass is given an initial displacement of 0.001 m while the right-hand mass is held in equilibrium and the system is released from rest.

Given: 0 Find:

0,

100

,

0

0,

158.11 0

1 , 0.5

1 , 2

,

0.

0

0.001 m,

,

Solution: The general solution of the differential equations is 1 cos 158.11 0.5

1 1 cos 100 sin 100 2 2 1 sin 158.11 0.5

Application of initial conditions to Eq.(a) leads to 0.001 2

0.5

100

151.1

2 100

The solution of Eqs. (c) through (e) is leads to

0 0

0.5 158.1

0.0002,

0

0,

0.0008,

0.0002 cos 100

0.0008 cos 158.1

0.0004 cos 100

0.0004 cos 158.1

0 which

Problem 6.17 illustrates the free response of a 2DOF system.

6.18 Determine the response of the system of Figure P6.2 and Chapter Problems 6.2 and 6.11 if the particle is given an initial velocity of 2 m/s when the system is in equilibrium.

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Chapter 6: Two Degree-of-Freedom Systems 16.69

Given:

1 , 1.128 0

0,

Find:

,

0

,

36.69

,

1 , 0.2949 0,

0

0,

0

2 m/s

Solution: The general solution of the differential equations is 1 cos 16.69 1.128

1 sin 16.69 1.128

1 cos 36.69 0.2949

1 sin 36.69 0.2949

Application of initial conditions to Eq.(a) leads to 0 1.128

0.2949

0

16.69

36.69

0

1.128 16.69

The solution of Eqs. (c) through (e) is leads to

0.2949 36.69

0,

0

0.0842,

0,

0.0842 cos 16.69

0.0383 cos 36.69

0.0950 cos 16.69

0.0113 cos 36.69

0.0383 which

Problem 6.18 illustrates the free response of a 2DOF system.

6.19 Determine the response of the system of Figure P6.4 and Chapter Problems 6.4 and 6.13 if the right-hand disk is given an angular displacement of 2° clockwise from equilibrium and the left-hand disk is given an angular displacement of 2° counterclockwise.

445 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems

Given: 0

2°,

Find:

,

64.1

,

0

0,

164.6 0

1 , 2.106 0.

,

2°,

0

1 , 0.358

Solution: The general solution of the differential equations is 1 cos 64.1 2.106

1 sin 64.1 2.106

1 cos 164.6 0.358

1 sin 164.6 0.358

Application of initial conditions to Eq.(a) leads to 2 2.106

0.358

2

64.1

164.1

0

0.358 164.6

2.106 64.1

0

0.5207,

The solution of Eqs. (c) through (e) is leads to

0,

0.5207 cos 64.1

2.5207 cos 164.6

1.0976 cos 64.1

0.0924 cos 36.69

2.5207

0 which

Problem 6.19 illustrates the free response of a 2DOF system.

6.20 Determine the response of the system of Chapter Problem 6.14 if the crane is disturbed resulting in an initial velocity of 10 m/s downward.

Given: 10 m/s

0

Find:

,

0,

0

0,

0

0,

0

Solution: The natural frequencies determined in the solution of Chapter Problem 6.14 are 446 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems 266.6

rad s

1513.7

rad s

The modal fractions are determined from 1000

6.16 5.24 10

10

5.24 300

10 5.24 10

1

0 0

Using the first equation gives 1000 6.16 5.24 10

10

leading to 1 1.0309

1 3.1971

The general solution of the differential equations is 1 1 cos 266.6 sin 266.6 1.0399 1.0309 1 sin 1513 3.171

1 cos 1513 3.171

Application of the initial conditions gives

0

0

0

0

0

1.0399

3.171

0

0

0

266.6

1513

0

266.6 1.0399

1513

3.171

0,

10

,

0,

1.57

10

1 sin 1513 3.171

10

The solution of the equations is resulting in 8.91

10

0

8.91

1 sin 266.6 1.0309

10

1.57

10

Problem 6.20 illustrates application of initial conditions to the response of a 2DOF system.

6.21 Determine the output from the seismometer of Chapter Problem 6.15 if the 60 g mass is given an initial velocity of 15 m/s. Use a two degree-of-freedom system, remembering that the seismometer records the relative displacement between the seismic mass and the body whose vibrations are to be measured.

Given:

0

0,

0

0,

0

15,

0

0

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Chapter 6: Two Degree-of-Freedom Systems Find: Solution: The natural frequencies are determined in the solution of Chapter Problem 6.15 as 35.95 , 152.3 . The modal fractions are determined from the equations 0.06

1390 40

1

40 40

0.03

0 0

which leads to 0.06

1390 40

35.95

Equation (b) is evaluated for 152.3

32.81 and for

, leading to

0.0429. Thus the mode shape vectors are

leading to 1 32.81

1 0.0429

The general solution of the differential equations is 1 1 cos 35.95 sin 35.95 32.81 32.81 1 sin 152.3 0.0429

1 cos 152.3 0.0429

Application of the initial conditions to Eq. (d) leads to 0 32.81

0.0429

0

35.95

152.3

15

32.81 35.95 The solution of Eqs. (e) through (h) are response of the seismometer is

0.0429 152.3 0,

0

0.000545,

0,

0.0984. The

32.81 0.000545 sin 35.95 0.0429 0.0984 sin 152.3 0.000545 sin 35.95 0.0984 sin 152.3 0.0173 sin 35.85

0.1026 sin 153.2

Problem 6.21 illustrates (1) the application of initial conditions to the solution for a 2DOF system and (2) the effect of a seismometer on measuring the response of a system.

448 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems 6.22 Determine the free response of the system of Figure P6.6 if the left-hand mass is given an initial displacement of 0.001 m while the right-hand mass is held in equilibrium and the system is released from rest. Use m = 1 kg, k = 10,000 N/m and c = 100 N · s/m.

Given: m = 1 kg, k = 10,000 N/m, c = 100 N · s/m, 0 01, 0 0 Find:

0

0.001,

0

0,

,

Solution: The differential equations derived in the solution to Chapter Problem 6.6 are 2

3

3

0

and 0 Substituting given values into these equations leads to 2

300

100

30000

10000

0

100

100

10000

10000

0

Assuming a solution of the form 1 and substituting into the equations leads to 2

300

100

30000

10000

0

100

100

10000

10000

0

A non-trivial solution of the above equations exists if and only if 2 The solutions for

100 100

10000 10000

0

are

27.437 The value of

300 30000 100 10000

135.174 , 27.437

corresponding to each

135.174 , 32.291, 162.762

is given by

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Chapter 6: Two Degree-of-Freedom Systems 2

300 30000 100 10000

Substituting calculated values of into the equation for and recognizing that the first component of the mode shape vectors is equal to 1 leads to 27.437

135.174

27.437

135.174

1 0.4153

0.7035 1

0.4153

32.291

1 0.3542

162.762

1 2.2552

0.7035

The general solution of the differential equations is 1

.

0.4153 1 0.3542

.

0.7035

1 2.552

.

1 0.4153

.

0.7035

.

or .

1 0 cos 135.17 sin 135.17 0.4153 0.7035 1 0 sin 135.17 cos 135.17 0.4153 0.7035 1 1 . .

0.3542 2.552

Application of the initial conditions leads to 0 0 0 0

0

0 0

0.001

0.4153

0.7035

0.3542

2.2552

27.437

135.17

32.291

162.762

27.437 0.4153 135.17 0.7035 135.17 0.4153 27.437 0.7035 162.762 2.552

1.119 10 , Solution of these equations leads to 4.43 10 , 2.54 10 . The response of the system is

32.291 0.3542 4.41

10 ,

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Chapter 6: Two Degree-of-Freedom Systems 10

1 0 cos 135.17 sin 135.17 0.4153 0.7035 1 0 4.41 sin 135.17 cos 135.17 0.4153 0.7035 1 1 . .

4.43 2.54 0.3542 2.552 .

11.19

11.19 cos 135.17 7.7496 2.54 .

6.48

.

10

4.43 1.57

4.41 sin 135.17 6.0407 .

Problem 6.22 illustrates the damped response of a 2DOF system.

6.23 Determine the response of the system of Figure P6.7 if the lower mass is given a displacement from equilibrium of 0.004 m and the upper mass is held in its equilibrium position and the system is released. Use m = 5 kg, k = 4000 N/m and c = 30 N · s/m.

Given: m = 5 kg, k = 4000 N/m, c = 30 N · s/m, 0.004, 0 0, 0 0 Find:

0

0,

0

,

Solution: The differential equations derived in the solution of Chapter Problem 6.7 are 2

0

and 0

2 Substituting given values into these equations leads to 5

60

30

10

4000

0

30

0

30

Assuming a solution of the form 1 and substituting into the equations leads to 5

60

30

4000

0

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Chapter 6: Two Degree-of-Freedom Systems 10

30

30

0

A non-trivial solution of the above equations exists if and only if 5 The solutions for

60 4000 30

0

are 0, 33.9547, 4.0227

The value of

30 30

corresponding to each

26.2801 , 4.0227

26.2801

except 0 is given by 5

60 30

4000

The mode shape vector for the first mode is 0 1 Substituting calculated values of into the equation for component of the mode shape vectors is equal to 1 leads to 4.0227

26.28014

4.0227

26.28014

and recognizing that the first 1

0.5707

0.5774 1

0.5707

0.5774

1 7.5859

33.29547 The general solution of the differential equations is .

1 0.5707

1 0.5707 .

0.5774

0.5774 0 1

.

1 7.5859

.

or .

1 0 cos 26.2801 sin 26.2801 0.5707 0.5774 1 0 sin 26.2801 cos 26.2801 0.5707 0.5774 1 .

7.5859

0 1

Application of the initial conditions leads to 452 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems 0 0

0.004

0.5707

0.5774

0

4.0227

26.2801

0 0

0

0

7.5859 33.954

4.0227 0.5707 26.2801 0.5774 26.2801 0.5707 4.0227 0.5774 33.954 7.5859

3.785 10 , 8.57 10 , Solution of these equations leads to 1.027 10 , 2.154 10 . The response of the system is 10

1 0 cos 26.2801 sin 26.2801 0.5707 0.5774 1 0 0 8.57 sin 26.2801 cos 26.2801 10.27 0.5707 0.5774 1 1 .

2.154 7.5859 .

10

37.85

.

10.27

0 1

37.85 cos 26.2801 16.65 1 .

16.34

8.57 sin 26.2801 16.96

Problem 6.23 illustrates the damped response of a 2DOF system.

6.24 Determine the free response of the system of Figure P6.8 if the machine tool has initial velocities 5 rad/s if I = 0.03 kg of 0 0.8 m/s and 0 · m2, c = 100 N · m, m = 3 kg, a = 0.3 m, b = 0.4 m, and k = 3000 N/m.

Given: 0 0.8 m/s , 0 5 rad/s, I = 0.03 kg 2 · m , c = 100 N · m, m = 3 kg, a = 0.3 m, b = 0.4 m, k = 3000 N/m.

Find:

,

Solution: The differential equations are derived in the solution of Chapter Problem 6.9 as 0 0

Substituting in the given values leads to 453 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems 0.03 0.03

1.2 0.9

21 28

70 70

630 840

2100 2100

0 0

The free solution is obtained by assuming 1 which when substituted into the governing equations leads to 21 28

0.03 0.03

630 840

1.2 0.09

70 70

1 2100 2100

0 0

3.087

10

Setting the determinant to zero leads to 56.7

0.063

5131

2.058

10

0

whose roots are 803.17, 31.62, 32.26 The values of

for each value of

29.43

are determined from 0.03 1.2

21 70

630 2100

leading to 803.17

0.004328

31.62

0.003763

29.43

3.6828

32.26

0.7223

The general solution is 1 0.004328

1 .

0.003763 1 0 . cos 29.43 sin 29.43 3.6828 0.7223 1 0 sin 29.43 cos 29.43 3.6828 0.7223 .

Application of initial conditions leads to 0 0

0 0

0.004328 5

803.17

0

0 0.003763 31.62

3.6828 32.26

0.7223 29.43

0 5

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Chapter 6: Two Degree-of-Freedom Systems 0

0.8

3.4761

0.1190

10

7.638

140.43

85.08

0.8

The solutions are 6.203

10

7.237

1.435

10

10

which leads to 1 1 . .

7.638 0.004328 0.003763 1 0 .

1.435 cos 29.43 sin 29.43 3.6828 0.7223 1 0 7.237 sin 29.43 cos 29.43 3.6828 0.7223

10

6.203

10

6.203 0.0268

. .

7.237

7.368 .

0.0275 1.435 cos 29.43 5.2848 7.237 sin 29.43 26.652

0 sin 29.43 1.0365 0 cos 29.43 5.3865

Problem 6.24 illustrates the solution for the free response of a 2DOF system.

6.25 Determine the principal coordinates for the system of Figure P6.1 and Chapter Problem 6.10. Write the differential equations which the principal coordinates satisfy.

Given: System shown Find:

,

Solution: The natural frequencies and mode shape vectors are determined in the solution to 1 1 Chapter Problem 6.10 as 100 , 158.11 , and .

0.5 2 The principal coordinates are 1

1 0.5

2

0.5

1 2 0.5 2 The principal coordinates satisfy the differential equations 1

0.2

0.4

0.4

0.8

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Chapter 6: Two Degree-of-Freedom Systems 10,000

0

25,000

0

Problem 6.25 illustrates the calculation of the principal coordinates for a 2DOF system.

6.26 Determine the principal coordinates for the system of Figure P6.2 and Chapter Problem 6.11. Write the differential equations which the principal coordinates satisfy.

Given: system shown Find:

,

Solution: The natural frequencies and mode shape vectors are determined in the solution to Chapter Problem 6.11 as 16.69

,

36.69

,

coordinates are 1 0.2072 1

0.7027

1 , 1.128

1 . The principal 0.2949

1 0.2949 1.128 1 0.2949 1.128

0.2949

1.128

0.7027 1.605 The principal coordinates satisfy the differential equations 278.55

0

1346.15

0

Problem 6.26 illustrates the calculation of the principal coordinates for a 2DOF system.

6.27 Determine the principal coordinates for the system of Figure P6.3 and Chapter Problem 6.12. Write the differential equations which the principal coordinates satisfy.

Given: system shown Find:

, 456

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Chapter 6: Two Degree-of-Freedom Systems Solution: The natural frequencies and mode shape vectors are determined in the solution to Chapter Problem 6.12 as 2.219

,

7.943

,

coordinates are 1 0.1023 1

0.4914

1 , 1.827

1 . The principal 0.2082

1 0.2082 1.827 1 0.2082 1.827

0.2082

1.827

0.8977 0.4914 The principal coordinates satisfy the differential equations 4.924

0

63.09

0

Problem 6.27 illustrates the calculation of the principal coordinates for a 2DOF system.

6.28 Determine the principal coordinates for the system of Figure P6.4 and Chapter Problem 6.13.

Write the differential equations which the principal coordinates satisfy.

Given: system shown ,

Find:

Solution: The natural frequencies and mode shape vectors are determined in the solution to 1 1 Chapter Problem 6.13 as 61.11 , 164.6 , , .

2.106 0.358 The principal coordinates are 1

1 0.358 2.106 1

0.358

0.1453

1 0.358 2.106

2.106

0.4058

0.4058 0.8456 The principal coordinates satisfy the differential equations 3734.4

0

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Chapter 6: Two Degree-of-Freedom Systems 27093

0

Problem 6.28 illustrates the calculation of the principal coordinates for a 2DOF system.

6.29 Determine the principal coordinates for the system of Figure P6.8 if it had no damping. Write the differential equations which the principal coordinates satisfy. Use 0.03 kg · m , 3 kg, 0.3 m, 0.4 m, 3000 N/m Given: 0.03 kg · m , 0.4 m, 3000 N/m Find:

3 kg,

0.3 m,

,

Solution: The differential equations obtained in the solution of Chapter Problem 6.8 assuming no damping are 0 0 Substituting given values and writing the equations in matrix form leads to 0.0729 0.0471

0.0471 0.0428

2100 0

0 2100

0 0

The natural frequencies are calculated from 0.729 0.471

2100

0.471 0.428 2100

0.08984

2430

4.41

10

and are equal to 44.23

rad s

158.5

rad s

The modal fractions are obtained by 0.729 0.471

2100

1 0.471 0.428 2100

which leads to

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Chapter 6: Two Degree-of-Freedom Systems 0.729 0.471

2100

and 0.5888

1.3701

The principal coordinates satisfy 1 0.588 1.370

1 1

1.370

1.882

1 0.5888 1.3701

0.5888

1.376

1.376 0.7529 The principal coordinates satisfy the differential equations 1956

0

25092

0

Problem 6.29 illustrates principal coordinates for a 2DOF system.

6.30 Determine the principal coordinates for the system of Chapter Problem 6.9. Write the differential equations which the principal coordinates satisfy if I = 0.03 kg · m2, c = 0 N · s/m, m = 3 kg, a = 0.3 m, b = 0.4 m, and k = 3000 N/m.

Given: 0.4 m,

0.03 kg · m , 3000 N/m

Find:

,

3 kg,

0.3 m,

Solution: The differential equations obtained in the solution of Chapter Problem 6.9 assuming no damping are 0 0 Substitution of given values leads to 0.03 1.2 0.9 0.03

630 2100 840 2100

0 0

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Chapter 6: Two Degree-of-Freedom Systems The natural frequencies and mode shape vectors are determined in the solution to Chapter Problem 6.14 as 266.6 , 1513.7 . The modal fractions are determined from 1000ω 6.16 10 5.24 10 0

5.24 10 300ω 5.24 10

1

0 0

which leads to 1000ω 6.16 5.24 10 For 266.6 coordinates are

,

1.04. For

10

1513.7

1

1 3.20 1.04

1

1 3.20 1.04

3.20. The principal

,

3.20 1.04

0.928

0.290

0.290

0.301

The principal coordinates satisfy the differential equations 71076

0

2.913 10

0

Problem 6.30 illustrates the calculation of the principal coordinates for a 2DOF system.

6.31 Determine the response of the system of Figure P6.1 and Chapter Problem 6.10 due to a sinusoidal force 200 sin 110 N applied to the block whose displacement is using the method of undetermined coefficients.

Given:

= 200 N,

Find:

,

110 rad/s

Solution: The differential equations governing the system are derived in the solution of Chapter Problem 6.1 and modified to take the applied force into account. They are 10 0

0 20

10

2 1

1 3

200 sin 110 0 460

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Chapter 6: Two Degree-of-Freedom Systems The method of undetermined coefficients is used and a solution assumed as sin 110 Substitution of Eq. (b) into Eq. (a) leads to 110

10 2 1 10

10

1 10 110 20 3

200 0

10

The solution of Eq. (c) is 10

1.16 2.00

Problem 6.31 illustrates application of the method of undetermined coefficients applied to determine the steady-state response of a 2DOF system.

6.32 Determine the response of the system of Figure P6.1 and Chapter Problem 6.10 due to a sinusoidal force 200 sin 80 applied to the block whose displacement is using the Laplace transform method.

200 sin 80

Given:

Find: Response of system using Laplace transforms Solution: The differential equations governing the system are derived in the solution of Chapter Problem 6.1 and modified to take the applied force into account. They are 10 0

0 20

10

2 1

1 3

0 200 sin 80

Taking the Laplace transform of Eq. (a) assuming all initial conditions are zero leads to 10

0 200 80 64000

200000 100000 100000 20 300000

where

. Solving Eq. (b) leads to

200

5

10

1 5

10

64000 1.6

1.6 10

10 3.2

10

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Chapter 6: Two Degree-of-Freedom Systems Partial fraction decompositions yield and .

and

which can be inverted to determine

Problem 6.32 illustrates the use of the Laplace transform method to a two-degree-offreedom system.

6.33 Determine the response of the system of Figure P6.2 and Chapter Problem 6.11 due to a sinusoidal force 100 sin 70 N applied to the particle using the method of undetermined coefficients.

100 sin 70

Given: Find:

,

Solution: The differential equations are derived in the solution of Chapter Problem 6.2 and put in a matrix form in the solution of Chapter Problem 6.11. They are modified to take the applied force into account as 0.667 0 0 2

750 500

100 sin 70 0

500 1000

A steady-state solution is assumed as Θ

sin 70

Substitution of Eq. (b) into Eq. (a) leads to 0.667 70 500 Equation (c) is solved leading to

750

500 2 70

1000 Θ

0.0402, Θ

0.0023.

100 0

Problem 6.33 illustrates the use of the method of undetermined coefficients on 2DOF systems.

6.34 Determine the response of the system of Figure P6.2 and Chapter Problem 6.11 due to a sinusoidal moment 50 sin 90 N· m applied to the bar using the method of undetermined coefficients.

Given:

50 sin 90 462

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Chapter 6: Two Degree-of-Freedom Systems Find:

,

Solution: The differential equations are derived in the solution of Chapter Problem 6.2 and put in a matrix form in the solution of Chapter Problem 6.11. They are modified to take the applied force into account as 0.667 0 0 2

750 500

0 50 sin 90

500 1000

A steady-state solution is assumed as Θ

sin 90

Substitution of Eq. (b) into Eq. (a) leads to 0.667 90 500

750

500 2 90

3.55

Equation (c) is solved leading to

1000 Θ

0 50

3.30

10 .

10 , Θ

Problem 6.34 illustrates the use of the method of undetermined coefficients on 2DOF systems.

6.35 Determine the response of the system of Figure P6.1 and Chapter Problem 6.10 due to a (a) a unit impulse applied to the block whose displacement is , and (b) a unit impulse applied to the block whose displacement is .

Given: Find:

applied first to block 1 and then to block 2 ,

Solution: The differential equations governing the motion of the system are derived in the solution of Chapter Problem 6.1 as 10 0 0 20

10

2 1

1 3

Taking the Laplace transform of the equations assuming all initial conditions are zero leads to 10

200000 100000 100000 20 300000 463

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Chapter 6: Two Degree-of-Freedom Systems where

and 200

(a) Setting

5

Solving Eq. (b) leads to

1 10

5

1 10 1

20 10

300000 200000

300000 10 5

10

10 10

10

10 10

0 leads to the transfer functions ,

20 5

200

1 ,

200

5

5

The response due to a unit impulse applied to the particle whose displacement is is the and , . Partial fraction decompositions yield inverse of the transfer functions , 0.03811 23596

0.0619 1043

0.0011275 23596

0.0011275 1043

,

,

The impulsive responses are 0.248 sin 153.61

1.917 sin 32.296 mm

7.314 sin 153.61 (b) Setting

34.9 sin 32.296

m

0 in the transfer functions leads to 1 ,

,

200 200

10 10

5

200000 10 5

10

5

10 5

10

A partial fraction decomposition yields ,

0.0079 23596

0.0402 1043

Inverting the transfer functions leads to the impulsive responses 7.314 sin 153.61

34.9 sin 32.296

m

7.9 sin 153.61

40.2 sin 32.296 mm

Problem 6.35 illustrates the transfer functions for a 2DOF system and the system’s impulsive responses.

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Chapter 6: Two Degree-of-Freedom Systems

6.36 Determine the response of the system of Figure P6.1 and Chapter Problem 6.10 due to the force of Figure P6.36 applied to the block whose displacement is .

Given: force shown applied to block whose displacement is Find:

,

Solution: The differential equations governing the motion of the system are derived in the solution of Chapter Problem 6.1 as 10 0 0 20

2 1

10

1 3

0

Taking the Laplace transform of the equations assuming all initial conditions are zero leads to 10 where

200000 100000 100000 20 300000 and F

0

Solving Eq. (b) leads to

200

5

1 10

20 5

10

300000 10

1

This leads to the transfer functions ,

300000 10 5

10

5

10 10

10

200

10000

20 5

200

1 ,

200

5

The force is 100 100

0.01 10000

0.01

0.01

0.01

100000

0.02

0.02 0.02

The Laplace transform is obtained by the second shifting theorem 100

10000

.

10000

.

The Laplace transforms of the responses are obtained by multiplying the transfer functions

by F(s) 465 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Two Degree-of-Freedom Systems ,

1.0482

100 20 300000 200 5 10 5 10 10000 20 300000 200 5 10 5 10

0.1386 153.61 1.0482 100 ,

200 0.0180 153.61 0.377 100

10 5 10 1 10 5 10

1.1868 32.73

.

.

.

1

200

0.377

1.1868 32.73 0.1386 153.61

.

0.3956 32.73 0.0180 153.61

5 5

10 10

.

0.3956 32.73

.

.

.

Inverting the transforms using the second shifting theorem yields 1.082

0.377

0.1386 cos 153.61 1.1868 cos 32.73 100 1.0482 0.01 9.02 10 sin 153.61 0.0110 sin 32.73 0.01 0.01 100 1.0482 0.02 9.02 10 sin 153.61 0.0110 sin 32.73 0.02 0.02 0.0180 cos 153.61 0.3956 cos 32.73 100 0.377 0.01 1.149 10 sin 153.61 0.0121 sin 32.73 0.01 0.01 100 0.377 0.02 1.149 10 sin 153.61 0.0121 sin 32.73 0.02 0.02

0.01 0.02

0.01 0.02

Problem 6.36 illustrates the use of transfer functions to determine the transient response of a 2DOF system.

6.37 Determine the response of the system of Figure P6.2 and Chapter Problem 6.11 due to a unit impulse applied to the particle.

Given: system shown 466 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.